Answer to Question #259590 in Calculus for Alunsina

Question

Find the domain and range and sketch the graph of the function f(x)= sqrt 16+6x-x²

Answer 1

Let us find the domain and range of the function $f(x)= \sqrt {16+6x-x^2}.$

Taking into account that

$f(x)= \sqrt {16+6x-x^2}=f(x)= \sqrt {25-(9-6x+x^2)}= \sqrt {25-(3-x)^2},$

we conclude domain contains all points $x$ such that $25-(3-x)^2\ge0,$ which is equivalent to $(3-x)^2\le 25.$ It follows that $|x-3|\le 5,$ and hence $-5\le x-3\le 5.$ We conclude that $-2\le x\le 8,$ and consequently, the domain of $f$ is $[-2,8].$ It follows from definition of $f(x)= \sqrt {25-(3-x)^2}$ that the maximum of $f$ is at the point $x=3,$ that is the maximum is equal to $5.$ The minimum is at the point $x=8.$ Therefore, the range of $f$ is $[0,5].$

Let us sketch the graph of the function $f(x)= \sqrt {16+6x-x^2}:$